Problem: Divide the following complex numbers. $ \dfrac{-10+10i}{1-3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1+3i}$ $ \dfrac{-10+10i}{1-3i} = \dfrac{-10+10i}{1-3i} \cdot \dfrac{{1+3i}}{{1+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-10+10i) \cdot (1+3i)} {(1-3i) \cdot (1+3i)} = \dfrac{(-10+10i) \cdot (1+3i)} {1^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-10+10i) \cdot (1+3i)} {(1)^2 - (-3i)^2} = $ $ \dfrac{(-10+10i) \cdot (1+3i)} {1 + 9} = $ $ \dfrac{(-10+10i) \cdot (1+3i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-10+10i}) \cdot ({1+3i})} {10} = $ $ \dfrac{{-10} \cdot {1} + {10} \cdot {1 i} + {-10} \cdot {3 i} + {10} \cdot {3 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{-10 + 10i - 30i + 30 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-10 + 10i - 30i - 30} {10} = \dfrac{-40 - 20i} {10} = -4-2i $